Although it is a question related to physics, since the point it really matters is its mathematical aspect, I post this question on MSE.

There's an additional exercise from Introduction to Electrodynamics by Griffith.

Problem 4.34 A point dipole p is imbedded at the center of a sphere of linear dielectric material (with radius $R$ and dielectric constant $K$). Find the electric potential inside and outside the sphere.

Now this problem can be solved by using separation of variables method, but I want to develop somewhat straightforward method other than separation of variables. (I do not merely want to solve this problem; I want to check if there can be other understanding from this problem)

So my idea is like this: As the medium is linear dielectric, $$P=\epsilon_0(K-1)E$$ and $$E=E_p + E_D$$ where $E_p$ is electric field by the point dipole and $E_D$ is by polarized dielectric.

Now there's no net charge other than the point dipole and the surface charge of the sphere, let's consider the surface charge density of the sphere. $$\sigma(r')=P \cdot \hat{r'} $$ $$=\epsilon_0(K-1)E \cdot \hat{r'}$$ $$=\epsilon_0(K-1)(E_p + E_D) \cdot \hat{r'}$$ but since $E_p$ is given as $$E_p (r') = {1 \over 4\pi \epsilon_0} {[3(p \cdot \hat{r'})\hat{r'}-p]\over r'^3},$$ $$E_p (r') \cdot \hat{r'} = {1 \over 4\pi \epsilon_0} {2p \cdot \hat{r'}\over R^3}$$ $$={2V_p(r')\over R}$$ where $V_p$ is electric potential by the point dipole. So $\sigma$ can be rewritten as $$\sigma(r')=\epsilon_0(K-1)(2V_p(r')/R + E_D \cdot \hat{r'}) $$. So now we can calcaulate electric potential $V_D$ by dielectric. $$V_D={1 \over 4\pi \epsilon_0}\oint_S{\sigma dA'\over \eta}$$ (where $\eta$ is separation vector given as $\eta = r - r'$ (or its magnitude by context)) $$={\epsilon_0(K-1) \over 4\pi \epsilon_0}\oint_S{(2V_p(r')/R + E_D \cdot \hat{r'}) dA'\over \eta}$$ Now let's focus on the second term of the integrand. $$\oint_S {E_D \cdot \hat{r'}\over \eta} dA' = \oint_S {E_D \over \eta} \cdot (\hat{r'} dA') $$ $$=\oint_S {E_D \over \eta} \cdot d\vec{A'}$$ (which means the scalar $dA'$ is now turned into the vector $d\vec{A'}$. Let me omit the vector notation for convenience after this.) But since $E_D=-\nabla' V_D$, $$\oint_S {E_D \over \eta} \cdot d\vec{A'}=-\oint_S {\nabla' V_D \over \eta} \cdot d\vec{A'}$$ (By Gauss's theorem,) $$=-\int_V \nabla' \cdot {\nabla' V_D \over \eta} d\tau'$$ (By the formula $\nabla fg = f \nabla g + g \nabla f, \nabla'^2V_D = 0$ in $V$) $$=-\int_V \nabla' ({1 \over \eta}) \cdot (\nabla' V_D ) d\tau'$$ (Since $(\nabla f) \cdot A = -\nabla \cdot (fA) + f \nabla \cdot A$ ,) $$=\int_V V_D \nabla'^2 ({1 \over \eta}) d\tau' - \oint_S V_D \nabla'({1 \over \eta}) \cdot dA'$$ But as we know that $$\nabla'^2({1\over \eta})=-4\pi\delta^3(\eta),$$ $$\int_V V_D \nabla'^2 ({1 \over \eta}) d\tau'=-4\pi\int_V V_D \delta^3(\eta) d\tau'$$ $$=-4\pi V_D(r) $$(if r is in V) $$=0 $$(if r is out of V)

And the second term is $$\oint_S V_D \nabla'({1 \over \eta}) \cdot dA' = \oint_S V_D ({\hat{\eta} \over \eta^2}) \cdot dA' = \oint_S V_D d\Omega' = 4\pi\bar{V_D}$$ which means that the integral is $(4\pi)\cdot$(Average of $V_D$ over solid angle measured from $r$) So $V_D$, the potential that we wanted to find, is $$V_D={(K-1) \over 4\pi }\oint_S{(2V_p(r')/R) dA'\over \eta}+(K-1)(- V_D(r)+{\bar{V_D}})$$ in the case where $r$ is in V. In turn, $$K \cdot V_D(r) - (K-1) {\bar{V_D}}(r)={(K-1) \over 4\pi R }\oint_S{2V_p(r') dA'\over \eta}$$

Now, this is the point where I'm stuck. My problem will be solved if:

$${\bar{V_D}}(r) = {2\over 3}V_D(r)$$ $$ \oint_S{2V_p(r') dA'\over \eta}={2\over 3}{r^3\over R^2}V_p(r)$$

Now, to make this question not be restricted to this one specific problem, I extract two main questions;

  1. Does the average of harmonic function over solid anlge $\bar{V_D}(r)$ have meaning (in other words, can it be expressed in the closed form?)

  2. Can we evaluate $${4\pi\epsilon_0\over p}\oint_S{V_p(r') dA'\over \eta}=\int_{0}^{2\pi}\int_{0}^{\pi} {\cos\phi \sin\phi \over \sqrt{R^2+r^2-2Rr(\cos\phi \cos\theta+\sin\phi \sin\theta \cos\psi )}} d\phi\ d\psi$$ (Although I asked this integral question in this post(Integral evaluation), it turned out that it might be hard to evaluate it. Since now the context is given, one might come up with easier form of this integral so that it can be evaluated.)

  3. If this approach is invalid, can there be other method that can substitute separation of variables for this kind of boundary value problems?

Any kind of opinions, hints, advices or answers will be appreciated! Please ask me back if there's ambiguity in my question.


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